Lesson A A vector quantity is a quantity that is fully described by both magnitude and direction. A scalar quantity is a quantity that is fully described by its magnitude. Examples of vector quantities that have been previously discussed include displacement, velocity, acceleration, and force. Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Vector diagrams were introduced and used in earlier units to depict the forces acting upon an object. Such diagrams are commonly called as free-body diagrams. Vectors can be directed due East, due West, due South, and due North. But some vectors are directed northeast (at a 45 degree angle); and some vectors are even directed northeast, yet more north than east. The direction of a vector is often expressed as an angle of rotation of the vector about its "tail" from east, west, north, or south. The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "tail" from due East. The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale. Directions are described by the use of some convention. The most common convention is that the direction of a vector is the counterclockwise angle of rotation which that vector makes with respect to due East. Lesson B Two vectors can be added together to determine the result (or resultant). For example, a vector directed up and to the right will be added to a vector directed up and to the left. There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors. The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors that make a right angle to each other. The method is not applicable for adding more than two vectors or for adding vectors that are not at 90-degrees to each other. The Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle.The direction of a resultant vector can often be determined by use of trigonometric functions. The head-to-tail method is employed to determine the vector sum or resultant. The head-to-tail method involves drawing a vector to scale on a sheet of paper beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins (thus, head-to-tail method). The process is repeated for all vectors that are being added. Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to real units using the given scale. The direction of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East. A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.
Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
Pick a starting location and draw the first vector to scale in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
Starting from where the head of the first vector ends, draw the second vector to scale in the indicated direction. Label the magnitude and direction of this vector on the diagram.
Repeat steps 2 and 3 for all vectors that are to be added
Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as Resultant or simply R.
Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
Measure the direction of the resultant using the counterclockwise convention discussed earlier in this lesson.
The problem involves the addition of three vectors: 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.SCALE: 1 cm = 5 m The head-to-tail method is employed as described above and the resultant is determined (drawn in red). Its magnitude and direction is labeled on the diagram. SCALE: 1 cm = 5 m Interestingly enough, the order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction. For example, consider the addition of the same three vectors in a different order. 15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.SCALE: 1 cm = 5 m When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (20. m, 312 degrees). The order in which vectors are added using the head-to-tail method is insignificant.SCALE: 1 cm = 5 m
Method 1: Lesson C and D
The resultant is the vector sum of two or more vectors. If displacement vectors A, B, and C are added together, the result will be vector R. When displacement vectors are added, the result is a resultant displacement. Any two vectors can be added as long as they are the same vector quantity. If two or more velocity vectors are added, then the result is a resultant velocity. The resultant vector (whether a displacement vector, force vector, velocity vector, etc.) is the result of adding the individual vectors. A vector is a quantity that has both magnitude and direction. Any vector directed in two dimensions can be thought of as having an influence in two different directions. That is, it can be thought of as having two parts. Each part of a two-dimensional vector is known as a component. The components of a vector depict the influence of that vector in a given direction. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector. The single two-dimensional vector could be replaced by the two components. Consider an airplane that is flying from Chicago's O'Hare International Airport to a destination in Canada. Suppose that the plane is flying in such a manner that its resulting displacement vector is northwest. If this is the case, then the displacement of the plane has two components - a component in the northward direction and a component in the westward direction. This is to say that the plane would have the same displacement if it were to take the trip into Canada in two segments - one directed due North and the other directed due West. If the single displacement vector were replaced by these two individual displacement vectors, then the passengers in the plane would end up in the same final position. The combined influence of the two components is equivalent to the influence of the single two-dimensional displacement.
Method 1: Lesson E
Any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components). That is, any vector directed in two dimensions can be thought of as having two components. For example, if a chain pulls upward at an angle on the collar of a dog, then there is a tension force directed in two dimensions. This tension force has two components: an upward component and a rightward component. The process of determining the magnitude of a vector is known as vector resolution.
The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. This method involves drawing the vector to scale in the indicated direction, sketching a parallelogram around the vector such that the vector is the diagonal of the parallelogram, and determining the magnitude of the components (the sides of the parallelogram) using the scale. If one desires to determine the components as directed along the traditional x- and y-coordinate axes, then the parallelogram is a rectangle with sides that stretch vertically and horizontally. A step-by-step procedure for using the parallelogram method of vector resolution is:
Select a scale and accurately draw the vector to scale in the indicated direction.
Sketch a parallelogram around the vector: beginning at the tail of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the head of the vector; the sketched lines will meet to form a rectangle (a special case of a parallelogram).
Draw the components of the vector. The components are the sides of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward velocity component might be labeled vx; etc.
Measure the length of the sides of the parallelogram and use the scale to determine the magnitude of the components in real units. Label the magnitude on the diagram
The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. The method of employing trigonometric functions to determine the components of a vector are as follows:
Construct a rough sketch (no scale needed) of the vector in the indicated direction. Label its magnitude and the angle that it makes with the horizontal.
Draw a rectangle about the vector such that the vector is the diagonal of the rectangle. Beginning at the tail of the vector, sketch vertical and horizontal lines. Then sketch horizontal and vertical lines at the head of the vector. The sketched lines will meet to form a rectangle.
Draw the components of the vector. The components are the sides of the rectangle. The tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward force velocity component might be labeled vx; etc.
To determine the length of the side opposite the indicated angle, use the sine function. Substitute the magnitude of the vector for the length of the hypotenuse. Use some algebra to solve the equation for the length of the side opposite the indicated angle.
Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.
Method 1: Lesson 1 G and H
Lesson G A motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. Motion is relative to the observer. A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. The resulting velocity of the plane is the vector sum of the two individual velocities. Since the plane velocity and the wind velocity form a right triangle when added together in head-to-tail fashion, the angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions. The resultant velocity of the motorboat can be determined in the same manner as was done for the plane. Motorboat problems such as these are typically accompanied by three separate questions: a) What is the resultant velocity (both magnitude and direction) of the boat? B)If the width of the river is X meters wide, then how much time does it take the boat to travel shore to shore? C) What distance downstream does the boat reach the opposite shore? The motion of the riverboat can be divided into two simultaneous parts - a motion in the direction straight across the river and a motion in the downstream direction. The boat's motor is what carries the boat across the river the Distance A; and so any calculation involving the Distance A must involve the speed value labeled as Speed A (the boat speed relative to the water). Similarly, it is the current of the river that carries the boat downstream for the Distance B; and so any calculation involving the Distance B must involve the speed value labeled as Speed B (the river speed. It is often said that "perpendicular components of motion are independent of each other." As applied to riverboat problems, this would mean that an across-the-river variable would be independent of a downstream variable. Lesson H A force vector that is directed upward and rightward has two parts - an upward part and a rightward part. That is to say, if you pull upon an object in an upward and rightward direction, then you are exerting an influence upon the object in two separate directions - an upward direction and a rightward direction. These two parts of the two-dimensional vector are referred to as components. A component describes the affect of a single vector in a given direction. AThe vector sum of these two components is always equal to the force at the given angle. The two perpendicular parts or components of a vector are independent of each other. A change in the horizontal component does not affect the vertical component. Changing a component will affect the motion in that specific direction. While the change in one of the components will alter the magnitude of the resulting force, it does not alter the magnitude of the other component.
Method 3: Lesson 2 A and B
Lesson A1. What is a projectile?- A projectile is an object that has only gravity acting upon it; if there are other forces acting upon it, it is not a projectile.2. How many forces act upon projectiles?- One; gravity3. What are some examples of projectiles?- An object dropped from rest without air resistance
- An object thrown upward vertically without air resistance 4. Is a force required to keep an object in motion? Explain.- No, there doesn’t have to be a force to keep an object in motion. A force is only needed to keep an object accelerating, or it would move at constant speed.5. What would happen according to Newton’s first law if there were no gravity?- The object would move in a straight horizontal line at a constant speed, yet with gravity it would increasingly move downward. This is due to inertia.Central Idea/Theme: A projectile is an object with one force acting upon it: gravity. If there was no gravity, objects would continue motion at constant velocity because forces are only required for acceleration.Lesson B1. List two components of projectile motion.- Horizontal and vertical motion2. What is the acceleration of gravity?- Freefalling objects fall at an acceleration of 9.8 m/s3. Must there be a horizontal force for horizontal acceleration?- Yes; gravity acts perpendicular to the horizontal motion although perpendicular components of motion are independent, so object moves with constant horizontal velocity at downward acceleration.4. How would an object travel in constant motion at constant speed?- If there was no unbalanced force, or absence of gravity; object would travel in parabolic trajectory because the downward force of gravity accelerates them downward from an otherwise straight-line, gravity-free trajectory.5. How would a projectile cause a projectile to move?- Gravity causes the object to move downward vertically. Gravity doesn't affect horizontal displacement.Central Idea/Theme: Gravtiy does not affect the horizontal displacement of a projectile, only the vertical.
Activity: Vector Mapping
Group: Julia, Amanda, Ryan, Ben
We walked all around the cafeteria using the directions that the other group gave us so that we could find the total displacement.
Through calculating analytically, we found that the displacement was 22.02 m at 157.98 degrees.
Conclusion: This activity allowed us to compare the way that we find results through calculations to the actual experimenting itself. We found the percent error between what we should have got analytically and what we should have gotten graphically. We did find that the analytical percent error was larger, being 1.09%, versus the graphical percent error, which was 0.59%. This activity allowed us to see that all of the calculations that we have been doing do actually work when done physically, in real life. Some places of error could have been that the tape measurer wasn't completely straight, giving off slightly wrong values. Another source of error could be that we can only round to the nearest hundredth when measuring, while more precise and sturdy technology could be used.
Method 3: Lesson 2 C
Part A and B
1. What would a vector diagram look like showing a projectile?
It would show the x and y velocity vectors and their magnitudes labeled. Lengths of the vector arrows are representative of the magnitudes.
2. What happens to the horizontal velocity as the projectile moves?
Nothing happens to it because it stays constant throughout the trajectory, moving in inertia.
3. What happens to the vertical velocity vector?
It changes constantly, accelerating 9.8 m/s/s.
4. How would a projectile's x and y vectors change if it was launched horizontally?
The object would move at first with initial velocity until reaching the maximum height and then accelerate once again when going down, although the horizontal will stay constant.
5. What would a v-t graph show for this event?
6. What equation can be used to show vertical displacement of a projectile?
This can be shown through the formula of a free-falling object, which is y=0.5 x g x t2.
7. What do these variables stand for?
G is -9.8 m/s/s and t is time in seconds; this equation refers to object with no initial velocity.
8. How could horizontal displacement of a projectile be found?
Through the formula x=Vix x t.
9. What is the equation for vertical displacement for an angled-launched projectile?
y=Viy x t + 0.5 x g x t2. Viy is the initial vertical velocity in m/s, t is time in seconds, and g=-9.8 m/s/s
What is the position of a projectile launched at an angle to the horizontal? Central Idea/Theme: The velocity of the verticle accelerates at 9.8 m/s/s although the horizontal moves at constant speed. These circumstances are different though, in the case that gravity is present.
Activity: Ball in Cup Part I
Group: Julia, Amanda, Ryan, Ben Objectives:
- Measure the initial velocity of a ball
- Apply concepts from 2-D kinematics to predict the impact point of a ball in projectile motion
- Take into account trial-to-trial variations in the velocity measurement when calculating the impact point Pre Lab Questions
If you were to drop a ball, releasing it from rest, what information would be needed to predict how much time it would take for the ball to hit the floor? What assumption must you make?
I would need to know the height from the place the object fell to the ground. You must make the assumptions that there is a constant vertical acceleration of -9.8 m/s/s due to gravity, and that the initial velocity of the object is 0 m/s/s because before it was dropped, it was at rest.
If the ball in Question 1 is traveling at a known horizontal velocity when it starts to fall, explain how you would calculate how far it will travel before it hits the ground.
In all projectiles, in order for it to be considered a projectile, the horizontal velocity stays the same the entire time. Therefore, you can use this information to calculate how far it will travel before it hits the ground by plugging the horizontal velocity into distance=velocity/time, as well as the time.
A single Photogate can be used to accurately measure the time interval for an object to break the beam of the Photogate. If you wanted to know the velocity of the object, what additional information would you need?
In order to find the time interval, you would need to find the distance that the object traveled on the x-axis (horizontally), using distance and time to find the velocity.
Write your procedure and get approval from Ms. Burns before you proceed any further!
What data will you need to collect? Remember that you must run multiple trials. Keep in mind your end goal!
We need to find out what the height that the projectile will be dropping from is.
How will you analyze your results in terms of precision and/or in terms of accuracy?
Because we are able to calculate so many different aspects of the projectile through different formulas specific to them, we are able to find theoretical values. We can then use these theoretical values and compare them to experimental values that we find through our experimenting. We can use the percent error formula, and the percent difference formula to analyze how accurate our results are.
Hypothesis: If our calculations are correct and we place the cup in the correct place, theoretically, then the ball should land in the cup. Available Materials:
Data Studio
meter stick or metric measuring tape
plumb bob
target
ramp
Carbon paper
masking tape
two right-angle clamps
Yellow Ball
newsprint
1Photogate Timer
Calipers
Procedure: The two videos below show how we fired the ball, aiming it at the angle of 25 degrees so that it would go into the cup.
Calculations:
We calculated the initial velocity using the components that we already knew in the equation.
This is taking into account the height of the cup, which we didn't initially account for. This calculation helped to decrease our percent error.
Conclusion: This activity was a sort of preview for the Shoot for Your Grade Lab. We were able to do all of the calculations to find different components necessary to getting the ball into the cup. Things that could have accounted for some error throughout the lab was the fact that at first, we didn't account for the height of the cup. When we did recalculate though, our percent error decreased even more than it from what it had been before. Other sources of error could have been from the fact that the ball does land in a slightly different place each time, as we saw through its placement on the paper in a triangular series of dots, never landing in the exact same place. Ways to fix this error would be to have a machine which shot the ball more precisely every time. Some error is eliminated because the cup does give some leeway as to where the ball lands exactly because its radius is larger than the ball. We were, after all, able to get the ball into the cup, and this shows that if we work hard to place the cup at the right placement, we should be able to get the ball into the cup every time, or close to it.
Activity: Ball in Cup Part 2
Group: Julia, Amanda, Ryan, Ben Objective: We want to place the hoops correctly so that the ball will go through all five of them and land into the cup, using theoretical calculations as a guide. Hypothesis: If our calculations are correct and we place all of the hoops in the correct places, theoretically, then the ball should go through all 5 hoops and land in the cup. Available Materials:
Data Studio
meter stick or metric measuring tape
plumb bob
target
ramp
Carbon paper
masking tape
two right-angle clamps
Yellow Ball
newsprint
1Photogate Timer
Calipers
Procedure: We put the ball into the medium speed of the shooter, set the shooter to 25 degrees, and shot the ball. We had previously hung the hoops on the ceiling by connecting them through the ceiling tiles and securing them with tape and calipers. We placed each hoop and the cup at specific coordinates, which we found through our calculations.
Calculations:
At the 25 degree angle, we were able to find the average distance that the ball would travel, using different components that we had taken down.
We found the initial velocity at the initial 25 degree angle to the horizontal.
These are the horizontal values that we used to place the five hoops at, and the last value is where the ball was supposed to land, so we placed that on the floor and that distance from the original shooting point was right in the middle of the cup, and we did account for the height of the cup.
These equations shown above show the vertical values for where the hoops needed to be placed so that the ball would go through them
This shows a diagram of the x and y coordinates (height and distance) from the launching point where we would place the hoops so that when the ball was shot at the 25 degree angle, it would make it through all five hoops and land in the cup.
Percent Error: Conclusion: We did calculate certain heights and distances at which we believed should be the center points of the hoops so that the ball would go through. These points included: (0.5, 0.168), (1.0, 0.210), (1.5, 0.125), (2.0, -0.088), (2.5, -0.433), and (2.953, -0.92). These heights and distances were all relative to the place from which the projectile was fired from and therefore, where it should have theoretically went. There was some error, as we found through our percent error calculations. This error ranged from 0.61% to 16.99% in the vertical values of the hoops. We had to end up guessing and checking to see where the hoop could be best positioned to match the path of the ball. We were able to get it through four hoops, still showing a significant amount of error, missing one hoop and the cup. Sources of error stemmed from the constant movement of the hoops between classes, which we couldn't control due to lack of space. Not only did we have to work with different approximations of placement, the tape that we used probably did give a little, slightly changing the position of the hoop. Due to some inconsistencies in the shooter and the angle, this could have contributed to error. We even saw on the paper how there were different distances that the ball traveled each time. This was probably caused by the angle being slightly differentiated each time. Different air pressures could have also slightly affected our results, considering that our calculations failed to account for them. We also were near an air vent, which caused some of our hoops to swing during our trials, slightly throwing off results. In order to decrease the amount of error, all groups should have had stationary individual places. Instead of using only tape, we should have use calipers to decrease movement of the strings and their give. Another way to decrease error would be to have a way of making sure that the shooter was at exactly 25 degrees each time, so that results wouldn't be affected by a changed angle. If we could measure air pressure with an instrument like a barometer, we could take it into account in our calculations to decrease percent error as well. It would also be helpful to stay in an area without any kind of ventilation nearby.
Gordorama Contest
Partner: Amanda Fava Vehicle Mass (g): 1.5 kg Time (s): 8.16 s Distance Traveled (m): 11.0 m Acceleration Value: -0.33 m/s2 Acceleration Calculation and Velocity Value: Pictures:
Improvements: It would have made our cart go faster if we were able to get the axels more aligned. Our car was constructed very well, although after the first time it crashed into the wall, the axels got slightly misaligned, and therefore, the car went into the side wall during the trials, rather than going straight. I think that was definitely the main problem because our wheels seemed to be working well and avoided as much friction as possible that would slow the vehicle down and cause it to eventually stop.
Method 1: Lesson 1 A and B
Table of Contents
Lesson A
A vector quantity is a quantity that is fully described by both magnitude and direction. A scalar quantity is a quantity that is fully described by its magnitude. Examples of vector quantities that have been previously discussed include displacement, velocity, acceleration, and force. Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Vector diagrams were introduced and used in earlier units to depict the forces acting upon an object. Such diagrams are commonly called as free-body diagrams. Vectors can be directed due East, due West, due South, and due North. But some vectors are directed northeast (at a 45 degree angle); and some vectors are even directed northeast, yet more north than east. The direction of a vector is often expressed as an angle of rotation of the vector about its "tail" from east, west, north, or south. The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "tail" from due East. The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale. Directions are described by the use of some convention. The most common convention is that the direction of a vector is the counterclockwise angle of rotation which that vector makes with respect to due East.
Lesson B
Two vectors can be added together to determine the result (or resultant). For example, a vector directed up and to the right will be added to a vector directed up and to the left. There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors. The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors that make a right angle to each other. The method is not applicable for adding more than two vectors or for adding vectors that are not at 90-degrees to each other. The Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle.The direction of a resultant vector can often be determined by use of trigonometric functions. The head-to-tail method is employed to determine the vector sum or resultant. The head-to-tail method involves drawing a vector to scale on a sheet of paper beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins (thus, head-to-tail method). The process is repeated for all vectors that are being added. Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to real units using the given scale. The direction of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East.
A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.
- Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
- Pick a starting location and draw the first vector to scale in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
- Starting from where the head of the first vector ends, draw the second vector to scale in the indicated direction. Label the magnitude and direction of this vector on the diagram.
- Repeat steps 2 and 3 for all vectors that are to be added
- Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as Resultant or simply R.
- Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
- Measure the direction of the resultant using the counterclockwise convention discussed earlier in this lesson.
The problem involves the addition of three vectors:20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.SCALE: 1 cm = 5 m
The head-to-tail method is employed as described above and the resultant is determined (drawn in red). Its magnitude and direction is labeled on the diagram.
SCALE: 1 cm = 5 m
Interestingly enough, the order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction. For example, consider the addition of the same three vectors in a different order.
15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.SCALE: 1 cm = 5 m When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (20. m, 312 degrees). The order in which vectors are added using the head-to-tail method is insignificant.SCALE: 1 cm = 5 m
Method 1: Lesson C and D
The resultant is the vector sum of two or more vectors. If displacement vectors A, B, and C are added together, the result will be vector R. When displacement vectors are added, the result is a resultant displacement. Any two vectors can be added as long as they are the same vector quantity. If two or more velocity vectors are added, then the result is a resultant velocity. The resultant vector (whether a displacement vector, force vector, velocity vector, etc.) is the result of adding the individual vectors.A vector is a quantity that has both magnitude and direction. Any vector directed in two dimensions can be thought of as having an influence in two different directions. That is, it can be thought of as having two parts. Each part of a two-dimensional vector is known as a component. The components of a vector depict the influence of that vector in a given direction. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector. The single two-dimensional vector could be replaced by the two components. Consider an airplane that is flying from Chicago's O'Hare International Airport to a destination in Canada. Suppose that the plane is flying in such a manner that its resulting displacement vector is northwest. If this is the case, then the displacement of the plane has two components - a component in the northward direction and a component in the westward direction. This is to say that the plane would have the same displacement if it were to take the trip into Canada in two segments - one directed due North and the other directed due West. If the single displacement vector were replaced by these two individual displacement vectors, then the passengers in the plane would end up in the same final position. The combined influence of the two components is equivalent to the influence of the single two-dimensional displacement.
Method 1: Lesson E
Any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components). That is, any vector directed in two dimensions can be thought of as having two components. For example, if a chain pulls upward at an angle on the collar of a dog, then there is a tension force directed in two dimensions. This tension force has two components: an upward component and a rightward component. The process of determining the magnitude of a vector is known as vector resolution.The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. This method involves drawing the vector to scale in the indicated direction, sketching a parallelogram around the vector such that the vector is the diagonal of the parallelogram, and determining the magnitude of the components (the sides of the parallelogram) using the scale. If one desires to determine the components as directed along the traditional x- and y-coordinate axes, then the parallelogram is a rectangle with sides that stretch vertically and horizontally. A step-by-step procedure for using the parallelogram method of vector resolution is:
- Select a scale and accurately draw the vector to scale in the indicated direction.
- Sketch a parallelogram around the vector: beginning at the tail of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the head of the vector; the sketched lines will meet to form a rectangle (a special case of a parallelogram).
- Draw the components of the vector. The components are the sides of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
- Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward velocity component might be labeled vx; etc.
- Measure the length of the sides of the parallelogram and use the scale to determine the magnitude of the components in real units. Label the magnitude on the diagram
The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. The method of employing trigonometric functions to determine the components of a vector are as follows:Method 1: Lesson 1 G and H
Lesson GA motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. Motion is relative to the observer. A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. The resulting velocity of the plane is the vector sum of the two individual velocities. Since the plane velocity and the wind velocity form a right triangle when added together in head-to-tail fashion, the angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions. The resultant velocity of the motorboat can be determined in the same manner as was done for the plane. Motorboat problems such as these are typically accompanied by three separate questions: a) What is the resultant velocity (both magnitude and direction) of the boat? B)If the width of the river is X meters wide, then how much time does it take the boat to travel shore to shore? C) What distance downstream does the boat reach the opposite shore? The motion of the riverboat can be divided into two simultaneous parts - a motion in the direction straight across the river and a motion in the downstream direction. The boat's motor is what carries the boat across the river the Distance A; and so any calculation involving the Distance A must involve the speed value labeled as Speed A (the boat speed relative to the water). Similarly, it is the current of the river that carries the boat downstream for the Distance B; and so any calculation involving the Distance B must involve the speed value labeled as Speed B (the river speed. It is often said that "perpendicular components of motion are independent of each other." As applied to riverboat problems, this would mean that an across-the-river variable would be independent of a downstream variable.
Lesson H
A force vector that is directed upward and rightward has two parts - an upward part and a rightward part. That is to say, if you pull upon an object in an upward and rightward direction, then you are exerting an influence upon the object in two separate directions - an upward direction and a rightward direction. These two parts of the two-dimensional vector are referred to as components. A component describes the affect of a single vector in a given direction. AThe vector sum of these two components is always equal to the force at the given angle. The two perpendicular parts or components of a vector are independent of each other. A change in the horizontal component does not affect the vertical component. Changing a component will affect the motion in that specific direction. While the change in one of the components will alter the magnitude of the resulting force, it does not alter the magnitude of the other component.
Method 3: Lesson 2 A and B
Lesson A1. What is a projectile?- A projectile is an object that has only gravity acting upon it; if there are other forces acting upon it, it is not a projectile.2. How many forces act upon projectiles?- One; gravity3. What are some examples of projectiles?- An object dropped from rest without air resistance- An object thrown upward vertically without air resistance
4. Is a force required to keep an object in motion? Explain.- No, there doesn’t have to be a force to keep an object in motion. A force is only needed to keep an object accelerating, or it would move at constant speed.5. What would happen according to Newton’s first law if there were no gravity?- The object would move in a straight horizontal line at a constant speed, yet with gravity it would increasingly move downward. This is due to inertia.Central Idea/Theme: A projectile is an object with one force acting upon it: gravity. If there was no gravity, objects would continue motion at constant velocity because forces are only required for acceleration.Lesson B1. List two components of projectile motion.- Horizontal and vertical motion2. What is the acceleration of gravity?- Freefalling objects fall at an acceleration of 9.8 m/s3. Must there be a horizontal force for horizontal acceleration?- Yes; gravity acts perpendicular to the horizontal motion although perpendicular components of motion are independent, so object moves with constant horizontal velocity at downward acceleration.4. How would an object travel in constant motion at constant speed?- If there was no unbalanced force, or absence of gravity; object would travel in parabolic trajectory because the downward force of gravity accelerates them downward from an otherwise straight-line, gravity-free trajectory.5. How would a projectile cause a projectile to move?- Gravity causes the object to move downward vertically. Gravity doesn't affect horizontal displacement.Central Idea/Theme: Gravtiy does not affect the horizontal displacement of a projectile, only the vertical.
Activity: Vector Mapping
Group: Julia, Amanda, Ryan, BenConclusion: This activity allowed us to compare the way that we find results through calculations to the actual experimenting itself. We found the percent error between what we should have got analytically and what we should have gotten graphically. We did find that the analytical percent error was larger, being 1.09%, versus the graphical percent error, which was 0.59%. This activity allowed us to see that all of the calculations that we have been doing do actually work when done physically, in real life. Some places of error could have been that the tape measurer wasn't completely straight, giving off slightly wrong values. Another source of error could be that we can only round to the nearest hundredth when measuring, while more precise and sturdy technology could be used.
Method 3: Lesson 2 C
Part A and B1. What would a vector diagram look like showing a projectile?
It would show the x and y velocity vectors and their magnitudes labeled. Lengths of the vector arrows are representative of the magnitudes.
2. What happens to the horizontal velocity as the projectile moves?
Nothing happens to it because it stays constant throughout the trajectory, moving in inertia.
3. What happens to the vertical velocity vector?
It changes constantly, accelerating 9.8 m/s/s.
4. How would a projectile's x and y vectors change if it was launched horizontally?
The object would move at first with initial velocity until reaching the maximum height and then accelerate once again when going down, although the horizontal will stay constant.
5. What would a v-t graph show for this event?
6. What equation can be used to show vertical displacement of a projectile?
This can be shown through the formula of a free-falling object, which is y=0.5 x g x t2.
7. What do these variables stand for?
G is -9.8 m/s/s and t is time in seconds; this equation refers to object with no initial velocity.
8. How could horizontal displacement of a projectile be found?
Through the formula x=Vix x t.
9. What is the equation for vertical displacement for an angled-launched projectile?
y=Viy x t + 0.5 x g x t2. Viy is the initial vertical velocity in m/s, t is time in seconds, and g=-9.8 m/s/s
What is the position of a projectile launched at an angle to the horizontal?
Central Idea/Theme: The velocity of the verticle accelerates at 9.8 m/s/s although the horizontal moves at constant speed. These circumstances are different though, in the case that gravity is present.
Activity: Ball in Cup Part I
Group: Julia, Amanda, Ryan, BenObjectives:
- Measure the initial velocity of a ball
- Apply concepts from 2-D kinematics to predict the impact point of a ball in projectile motion
- Take into account trial-to-trial variations in the velocity measurement when calculating the impact point
Pre Lab Questions
- If you were to drop a ball, releasing it from rest, what information would be needed to predict how much time it would take for the ball to hit the floor? What assumption must you make?
- I would need to know the height from the place the object fell to the ground. You must make the assumptions that there is a constant vertical acceleration of -9.8 m/s/s due to gravity, and that the initial velocity of the object is 0 m/s/s because before it was dropped, it was at rest.
- If the ball in Question 1 is traveling at a known horizontal velocity when it starts to fall, explain how you would calculate how far it will travel before it hits the ground.
- In all projectiles, in order for it to be considered a projectile, the horizontal velocity stays the same the entire time. Therefore, you can use this information to calculate how far it will travel before it hits the ground by plugging the horizontal velocity into distance=velocity/time, as well as the time.
- A single Photogate can be used to accurately measure the time interval for an object to break the beam of the Photogate. If you wanted to know the velocity of the object, what additional information would you need?
- In order to find the time interval, you would need to find the distance that the object traveled on the x-axis (horizontally), using distance and time to find the velocity.
- Write your procedure and get approval from Ms. Burns before you proceed any further!
- What data will you need to collect? Remember that you must run multiple trials. Keep in mind your end goal!
- We need to find out what the height that the projectile will be dropping from is.
- How will you analyze your results in terms of precision and/or in terms of accuracy?
- Because we are able to calculate so many different aspects of the projectile through different formulas specific to them, we are able to find theoretical values. We can then use these theoretical values and compare them to experimental values that we find through our experimenting. We can use the percent error formula, and the percent difference formula to analyze how accurate our results are.
Hypothesis: If our calculations are correct and we place the cup in the correct place, theoretically, then the ball should land in the cup.Available Materials:
Calculations:
Conclusion: This activity was a sort of preview for the Shoot for Your Grade Lab. We were able to do all of the calculations to find different components necessary to getting the ball into the cup. Things that could have accounted for some error throughout the lab was the fact that at first, we didn't account for the height of the cup. When we did recalculate though, our percent error decreased even more than it from what it had been before. Other sources of error could have been from the fact that the ball does land in a slightly different place each time, as we saw through its placement on the paper in a triangular series of dots, never landing in the exact same place. Ways to fix this error would be to have a machine which shot the ball more precisely every time. Some error is eliminated because the cup does give some leeway as to where the ball lands exactly because its radius is larger than the ball. We were, after all, able to get the ball into the cup, and this shows that if we work hard to place the cup at the right placement, we should be able to get the ball into the cup every time, or close to it.
Activity: Ball in Cup Part 2
Group: Julia, Amanda, Ryan, BenObjective: We want to place the hoops correctly so that the ball will go through all five of them and land into the cup, using theoretical calculations as a guide.
Hypothesis: If our calculations are correct and we place all of the hoops in the correct places, theoretically, then the ball should go through all 5 hoops and land in the cup.
Available Materials:
Calculations:
Percent Error:
Conclusion: We did calculate certain heights and distances at which we believed should be the center points of the hoops so that the ball would go through. These points included: (0.5, 0.168), (1.0, 0.210), (1.5, 0.125), (2.0, -0.088), (2.5, -0.433), and (2.953, -0.92). These heights and distances were all relative to the place from which the projectile was fired from and therefore, where it should have theoretically went. There was some error, as we found through our percent error calculations. This error ranged from 0.61% to 16.99% in the vertical values of the hoops. We had to end up guessing and checking to see where the hoop could be best positioned to match the path of the ball. We were able to get it through four hoops, still showing a significant amount of error, missing one hoop and the cup. Sources of error stemmed from the constant movement of the hoops between classes, which we couldn't control due to lack of space. Not only did we have to work with different approximations of placement, the tape that we used probably did give a little, slightly changing the position of the hoop. Due to some inconsistencies in the shooter and the angle, this could have contributed to error. We even saw on the paper how there were different distances that the ball traveled each time. This was probably caused by the angle being slightly differentiated each time. Different air pressures could have also slightly affected our results, considering that our calculations failed to account for them. We also were near an air vent, which caused some of our hoops to swing during our trials, slightly throwing off results. In order to decrease the amount of error, all groups should have had stationary individual places. Instead of using only tape, we should have use calipers to decrease movement of the strings and their give. Another way to decrease error would be to have a way of making sure that the shooter was at exactly 25 degrees each time, so that results wouldn't be affected by a changed angle. If we could measure air pressure with an instrument like a barometer, we could take it into account in our calculations to decrease percent error as well. It would also be helpful to stay in an area without any kind of ventilation nearby.
Gordorama Contest
Partner: Amanda FavaVehicle Mass (g): 1.5 kg
Time (s): 8.16 s
Distance Traveled (m): 11.0 m
Acceleration Value: -0.33 m/s2
Acceleration Calculation and Velocity Value:
Pictures:
Improvements: It would have made our cart go faster if we were able to get the axels more aligned. Our car was constructed very well, although after the first time it crashed into the wall, the axels got slightly misaligned, and therefore, the car went into the side wall during the trials, rather than going straight. I think that was definitely the main problem because our wheels seemed to be working well and avoided as much friction as possible that would slow the vehicle down and cause it to eventually stop.